Probabilistic Methods in Combinatorics: Homework ... - Google Sites

Probabilistic Methods in Combinatorics: Homework Assignment Number 1 Noga Alon

Solutions will be collected in class on Wednesday, March 25, 2009. n−1

1. Suppose n > 4 and let H be an n-uniform hypergraph with at most 4 3n edges. Prove that there is a coloring of the vertices of H by 4 colors so that in every edge all 4 colors are represented. 2. Prove that there is an absolute constant c > 0 with the following property. Let A be an n by n matrix with pairwise distinct entries. Then there is a permutation of the rows of A so that no √ column in the permuted matrix contains an increasing sub-sequence of length at least c n. 3. (i) Prove that every set A of n nonzero integers contains two disjoint subsets B1 , B2 ⊂ A, so that |B1 | + |B2 | > 2n/3 and each set Bi is sum-free (that is, there are no b1 , b2 , b3 ∈ Bi so that b1 + b2 = b3 .) (ii) Prove that the same conclusion holds for any set A of n nonzero reals. 4. Let {(Ai , Bi ), 1 ≤ i ≤ h} be a family of pairs of subsets of the set of integers such that |Ai |+|Bi | = k for all i, Ai ∩ Bi = ∅ and (Ai ∩ Bj ) ∪ (Aj ∩ Bi ) 6= ∅ for all i 6= j. Prove that h ≤ 2k . 5. Let X be a collection of pairwise orthogonal unit vectors in Rn and suppose the projection of each of these vectors on the first k coordinates is of Euclidean norm at least ǫ. Show that p |X| ≤ k/ǫ2 , and this is tight for all n = 2r , and ǫ = k/2r < 1, with r ≥ 1 an integer.

1 libxz - dwixehpianewa zeizexazqd zehiy 317610087 l`kin bxaxtiw 2009 uxna 25

1 dl`y

e

m`

xe

= 0-e

mirav 4-a dreav `l

e

xe

m`

e

xicbp

= 1

zyw lkl .z"ae cig` ote`a erav z` lixbp znev lkl -y xexa .mirav drax`a dreav

E [xe ] = 4


 n

 n

3

2

6

4

 n 1

+4

4

4

1 and every n-uniform hypergraph H with at most cn1/4 2n edges there is an ordering of the vertices of H such that there are no two edges A and B that intersect in a unique element, and all members of A − B precede all those of B − A, while the unique element in A ∩ B appears after all those of A − B and before all those of B − A. (ii) Apply (i) to conclude that for c, n and H as above, H is two-colorable. 2. (Every monotone property has a threshold). Let F be a family of graphs on n labeled vertices, and suppose F is monotone, that is, if F ∈ F and G contains all edges of F , then G ∈ F. Suppose ǫ > 0, 0 < p < 1 and suppose that the probability that the random graph G(n, p) ⌉}, the probability that G(n, q) ∈ F is at belongs to F is ǫ. Show that for q = min{1, p⌈ ln(1/ǫ) ǫ least 1 − ǫ. 3. Let v1 = (x1 , y1 ), . . . , vn = (xn , yn ) be n two dimensional vectors, where each xi and each yi is 2n/2 √ . Show that there are two disjoint nonempty sets a positive integer that does not exceed 100 n I, J ⊂ {1, 2, . . . , n} such that X X vi = vj . i∈I

j∈J

4. Call a family F of subsets of [n] = {1, 2, . . . , n} distinguishing if for every two distinct subsets A and B of [n] there is an F ∈ F so that |A ∩ F | = 6 |B ∩ F |. (i) Show that there exists such an F of size |F| ≤ (2 + o(1)) logn n . 3

(ii) Show that any such F is of size at least (2 − o(1)) logn n . 2

5. Show that there is a positive constant c such that the following holds. For any n vectors P a1 , a2 , . . . , an ∈ R2 satisfying ni=1 ||ai ||2 = 1 and ||ai || ≤ 1/10, where || · || denotes the usual Euclidean norm, if (ǫ1 , . . . , ǫn ) is a {−1, 1}-random vector obtained by choosing each ǫi randomly and independently with uniform distribution to be either −1 or 1, then P rob(||

n X

ǫi ai || ≤ 1/3) ≥ c.

i=1

6. The Haj´ os number of a graph G is the maximum number k such that there are k vertices in G
with a path between each pair so that all the k2 paths are internally pairwise vertex disjoint (and no vertex is an internal vertex of a path and an endpoint of another). Is there a graph whose chromatic number exceeds twice its Haj´os number ?

2 libxz - dwixehpianewa zeizexazqd zehiy 317610087 l`kin bxaxtiw 2009 lixt`a 22

zeillk zexrd

jXi j 6  i

.Pr [

.

;

(

= 1 2)]

>

3 :miiw 2 2



1

t
t
t
2t s 6 bt=2c miiw s lkl ok enk . bt=2c 6 pt+1

,

2 zepeye 0 zlgez ilra X1 ; X2 n"n bef xear :1 Berge htyn m

t > 3 xeare 22pm 6 2

m
22m m 6 p2m

2

miiw :ifkxn inepia mcwn zkxrd

` 1 dl`y

A \ B = fxg-y jk B -e A zezyw bef lkl .miznvd ly ixwn xeciq  idz :ik xexa ."B A-a miznvd lkl mcewy x-l ( itl) mincew A B -a miznvd lk" :rxe`nl XA;B

xehwicpi` dpzyn zeidl

XA;B = 1] =

Pr [

n (2n

[(

z` xicbp

2

1)!]

1)!

=

2

n

1

k
-n


nn [(

1

2

1)!]

(2

2)!

=

2

n

1


nn 2

1

2



1

6 n

1

2(

1

1)

p
n 2




1 2

n

2 2

n

3 2

6 pn 2

P

H -a zezyw k := cn1=4 2n -n xzei `l yi m` .X = XA;B xicbp k
23p 2n < c2 23 = 1 :zlgezd zeix`piln if` 3=2 xeciq xnelk ,X < 1 exear ,xeciq miiw okle c = 2 xear ,E [X ] 6 2 2 n ,

A; B

dl`k (mixecq) zebef 2

2

xzei `l okle

.yexck

a 1 dl`y

 xeciqd itl zyw idyefi`a oey`xd znevd `ed v m` legka ravi v znev .l"pk xeciq  idz A ly oexg`d xai`d hxta .legka dreav A ik xexa .zipeb-cg A gipp .mec`a .dxizqa ,XA;B = 1 f` la` .B ly

eravi miznvd x`y .

oey`xd xai`d `edy oeeikn legka reav

2 dl`y

 :=

dyrpy ieqipd lr xefgp . miieqipd



l

ln(1



=)

lky zexazqdd .minrt

m xear

 (F -l

q

=

p ik gipp okl

.1 zexaqzda

jiiy sxbd m`d dwicae

p

G(n; q) = Kn 2 F

zexazqda zezyw dxiga





q = 1 m` l"x) G(n; p)-a f`

p divwpetd ik al miyp .(1 6 exp ( ) 6 exp ln(1=)  =  lr dler dpi` elyki  0 zexazqdd xnelk ,1 (1 p) 6 p = q ik milawn .f (0) = 0 ,ok enk .[0; 1] rhwa dler dpi` f okle f 6 0 zniiwn dphw dpi`y zexazqda F -l jiiy epx`izy ieqipdn lawznd sxbd .q lr dler dpi` miieqipdn cg`a xgaiz zywy .1 -n dphw dpi`y zexazqda F -l jiiy ,dphw `l zyw zxigal zexazqdd eay ,G(n; q) okle ,1 -n

f (p) = 1

(1

p)

)

3 dl`y xexa .

Berge

V

 

=

itl

X = P  v ok enk ;Y = P  y -e X = P  x onqp .z"ae cig` ote`a  ; : : : ;  2 f0; 1g xgap 1 n i i i i i i i i i Y P 2 P P 1 1 2n 1 2n :dnec ote`a .V ar [X ] = .V ar [Y ] 6 i xi 6 4
1002 :oke E [Y ] = 2 i yi ,E [X ] = 2p i xi -y 4
1002 4 :( = 100 2 xear) h Pr

jX E X j; jY E Y j 6 [

]

[

]

2

(

n

=

1) 2

i

>

1

4




3 10

4

Berge, P.O (1937), A note on a form of Tchebyche's theorem for two variables. Biometrika 29:405-4061

1

2 1







1 1 4
103 4 -y o`kn lawzny jxr miiw okle ,mikxr 2 -n zegt milawn minekq 2
4 0 > 1-n xzei i"r P P 0 0 0 0 0 xexa .J = J K -e I = I K ,K = I \ J xicbp . 2 0 v = 2 0 v -y jk I 6= J  [n] zeniiw xnelk .minekq 3 4 10

n

n

i

j

0 did ynn miiaeig mixehwe ly mekq zxg`] I; J 6= ; ,I \ J = ; ik X X X X X X v = v v = v v = v i

I

j

J

:oke [

2

i

i

i

20

I

i

2

I

i

i

K

j

20

j

J

j

2

j

K

j

2

j

J

(zeillk zexrd) 4 dl`y

A (A)-a onqp .A ly A zecenrdn zakxeny dvixhnd z` A

log2  liaya lg  onqp

.

A -a

onqp

A  [n] dveawe k  n xcqn A dvixhn xear A ly zecenrd mekq z` (A) heyt e`)

:zelewy ze`ad zeprhd ik orhp .

A

jF \ Aj =6 jF \ B j zniiwnd F 2 F zniiw ,A 6= B  [n] lkly jk k lceba F  2[ ] zniiw  .(A) = 6 (B ) ,A =6 B  [n] lkly jk 0=1 ly k  n xcqn A dvixhn zniiw (*)   (A) = ik ze`xl lw .A = 6 B  [n] dpiidz .jtidle ,j 2 F m"m` A = 1 xicbp F = fF1 ; : : : ; F g xear ,ok`e .ipyd geqipd xear dl`yd z` ,ok m` ,xeztp .jF \ Aj = jF \ B j ,i lkl m"m` (A) = (B ) okle .jF \ Aj n

.

i

i

ij

i

k

i

i

` 4 dl`y

k) k  n xcqn (1=2 zexazqdae z"a ote`a 0 zeidl xgap da xai` lk xnelk) zixwn 0=1 zvixhn A idz X P = 1 m`y xexa ."(A) = (B )" :rxe`nd ly X xehwicpi` dpzyn xicbp A; B  [n] bef lkl .(jynda 0 0 .E [X ] := \ =; E [X ] lr dler dpi` miiwzn `l (*)-y zexazqddy o`kn .X 0 0 = 1-y jk zexf A ; B mb zeni rawii

-iw f`

A;B

A

A;B

A ;B

A;B

B

2

E [X

A;B

fp;qg X

min

] = 42

p

q

  

q `

p `

`=0

:ik xexa

3k 5

q = jB j-e p = jAj xy`k

:okle .

E [X ] =

2

n! 42 p ! q !( n p q)! 16 + 6 X

p

q

fp;qg X

min p

q

  

`=0

n

q `

p `

3k

X

=

5





n! 2 p!q!(n p q)!

p

q

:lawpe

n!

 k



1

 

 k+1

p+q p

k

t = p + q onqp

t n 2 2 st = s t ( n t )! s !( t s )! =0 =0 =1 =1 :f`e . = (n) = 2 + 1= lg lg n = 2 + o(1) xy`k k =  lg3

X n

E [X ] =

X t

X n

t

s

t

X t

tk

s

t

n

2

4n 2k

A(n) 6



n

lg n 2



n lg2 n n

:ixdy

n

2 lg2 n 2

lgnn

6 n 2e lg2 n


n2

lg n

2

E [X ] 6

P lgn2 n

A(n) =

lgnn

t




s

t=3

6

t

 




2

4n 2k

= o(1) 

2 (t + 1) ( 1) 2 = o(1)  2 + n lg2 n lg(2 lg ) = o(1)

b 2c t=

= o(1) ixdy

xgap

n

t

k

=

t

lgnn +lg n

=2

:o`kn

 

lgnn +2 lg(2n)

=2

n

e




n

t :miiw t > 3 lkl ik reci 6 p2+1 t

4n2 + X n 2 (t + 1) 2 + 6 4n2 + X n 2 6 k+1 2 t 2 t 2 (t + 1) (t + 1) k 2 1 =3 =3 n

tk

t

t

tk

k

k

t

2 6 42n +

n 2n   lg X n

k

= 2

n

t=3

k 1 (2 lg lg n

2



lg2 n 2 + t n t

lg n)+n lg 3

t

k 1

2

t=

+ o(1) 6 2 .yexck

n X

k

 

n

lg2 n

lg lg n+



lg2 n n 2 6 n t t

21 lg n

( 2 1)

n lg

3

 k21

=2

3 + o(1) = n

n lg 3 lg lg n +o( lg nlg n )

= o(1)

A dvixhn zniiw ziaeig zexazqda (lecb witqn n xear) okle 2

a 4 dl`y .z"ae cig` ote`a

i

i

1 2

okl . -e

i

Pr xen`dn ."si

2 f0; 1g xgap

1 ; : : : ; n

:= E [v ] = 12 s

si

= ([n])

vi

 n xcqn zizexixy dvixhn A idz = (A)-e A = fj j  = 1g xicbp

k

onqpe ,yexck

ik xexa .v

j

Cherno itl ,o`kn . 2 := V ar[v ] = 41 s -e

:



h vi

.s

mixhnxt mr zinepia bltzn

p

si

2 >

> pn log n mbe v

si



i

log s 6 2 exp

2s log s i

i



2 >

si

i



 i

6 2 exp ( 2 log s ) = 2s i

si

ps log s " (*) :rxe`nl xehwicpi` dpzyn zeidl X i

i

i

i

i

i

i

2

1 6 i 6 k lkl

xicbp

:lirl

2 (n log n) 1 = 2k(n log n) 1 6 log2 n = o(1) `l yi zeveawd xzi xear .(*) i`pzd z` miiwnd aikx likn (A) oxear A  [n] zeveaw o(2 )-n xzei `l yi okl p p :o`kn ,2 6 (2 n log n) + o(2 ) :dl`yd ii`pz z` zniiwn A-e li`ed .aikx lkl mipey mikxr 2 n log n-n xzei n(1 o(1)) k> > (2 o(1)) lg n(1 n+ o(1)) = (2 o(1)) lgnn lg +lg lg 1+ 2 [ ] := E

EX

hX

Xi

i

=

X

X

[ ]6

E Xi

n

k

n

n

n

n

5 dl`y   X

:mihpnen xtqn aygp .

E X2

[ ] =

2 3 X E4 xi xj i j 5

=

i;j



E X

4

X

=

X

[

xi xj E i j

]=

X

i;j

[

xi1 xi2 xi3 xi4 E i1 i2 i3 i4

]=

[

X

] =

x4i

i

:z` aygp .E

[X 2 + Y 2 ] = 1 hxtae ,Y [

xi1 xi2 yi3 yi4 E i1 i2 i3 i4

]=

i1 ;i2 ;i3 ;i4

X

=

Pn

=1  a = i

i

Pn

=1 

i

i

i

xi yi

onqp

x2i

i

X

i1 ;i2 ;i3 ;i4

E X 2Y 2

Y

 

 

+ 42

X

x2i x2j

i exp ( 2 ) :miiw . = xicbp 2 jxe`a lelqn xear f`e = 18 xicbp lkl P Y Y Y Y Y (1 )
= 2
= (1 )j ( )j > (1 ) = (1 ) = 1

ixefgn P " :rxe`nd zeidl AP xicbp P lelqn lkl . ; : : : ; r P

i

N P

Nj P

ij

j

Q

j

Nj P

Q

2 ( )

xQ

j

N P

= 18

i

e

Q

8

xP

N

Q

18

aj

ai

P

ij

9 > j

h

18 8 e

P

j

i

ai

Nj P

ij

aj

ai

i

aj

ai e

ij

j

j

P

j

i

Nj P

j

ij

ai

xQ

2 j( ) P

j

j

ai

xP

;

AQ

i

ai

xP

AP

AP

P

ij

e

AP ; AQ

2

j

i

i



> 18

16 

i

e

=

r

i

= Pr[ ] AP

.zraep dprhd (illkd gqepd) zilweld dnldn okle

4 dl`y ote`a da xag zeidl xgap znev lky jk miznv ly zixwn dveaw (ixlebx d-d sxbd `ed G xy`k) V e` Xv

= 0" :rxe`nd

Av

0 -e V -a v

ly mipkyd xtqn Xv eidi v

2

V

znev lkl .p

( ) \ ( ) = ; xy`k z"a

mr v xear miiwzn `l oexg`d oeieeyde li`ed .N u

N v

= (2 + 2ln )

d =d

Av -e Au

(

Pr [ ] 6 (1 ) + Pr( 3 ) 6 (1 ) + ( 2 27)2+2 ln 6 + .0 6 6 + 6 ln =: -y jk 0 dveaw zniiw zilweld dnld itl okle p

< Xv

d

X >

d

pd

k

p

V

1

d

e =

d

e

pd

zexazqdae z"a

3 1) 2 xzeid lkl 1 2 Pr[ ] 2 1 :miiw

zerxe`nd ik xexa ."Xv >

.u miznv d d Av

0  V [G] idz

e

pd

< d


2 Mi > Q2i=1 i

2k = 4

Mi xicbp ,1 6 6 2 i

2k

kX1 2k i=0

k

1 i

 =

k

znev lkl .1; : : : ; 2k :miznvd z` onqp

1 2

:K leitman itl .(zecxei) zeipehepen

Mi zepekzdy al miyp 6 dl`y

(V [G]; N mb ik xexa

E)

m"m` E

2 C -e (yxete) xiyw ( [ ] ) m"m` 2 A-y jk A C  2N dpiidz . = [ ] idz .ok jAj = 2jN j -y oezp .zcxei zipehepen C -e dler zipehepen A ik xexa .(yxete) xiyw jN j . 6 2 okle 22jN j = 2jN j jA \ Cj 6 jAjjCj = 22jN j 2 : itl .jCj = 2 V G ;E

jN j Q-e .jA \ Cj = 2

E

;

N

E G

P

Q

P

Q

P

2

K leitman

P

Probabilistic Methods in Combinatorics: Homework Assignment Number 4 Noga Alon

Solutions will be collected in class on Wednesday, June 3, 2009. 1. Let G1 , G2 , . . . , Gm be m graphs on the same set of vertices [n] = {1, 2, . . . , n}, and suppose that the chromatic number of each graph Gi is exactly k. Show that there is a partition of the set of vertices [n] into two disjoint sets A1 , A2 so that for every i, 1 ≤ i ≤ m, and for every j, 1 ≤ j ≤ 2, √ p the chromatic number of the induced subgraph of Gi on Aj is at least k/2 − 2 ln(2m) k. Hint: use an appropriate martingale to show that this holds with positive probability for a random partition. 2. Prove that there exists a positive constant δ > 0 and an integer n0 = n0 (δ) so that for all n > n0 and every collection S1 , S2 , . . . , Sm , where m ≤ 2δn , of subsets of [2n] = {1, 2, . . . , 2n}, satisfying |Si | = n for all i, there is a function f : [2n] 7→ [n] = {1, 2, . . . , n} so that for every i, 1 ≤ i ≤ m, 0.63n ≤ |f (Si )| ≤ 0.64n. Hint: 0.63 < 1 −

1 e

< 0.64.

3. Show that for any ǫ > 0 there is a C = C(ǫ) such that every set S of at least ǫ3n vectors in Z3n contains three vectors so that the Hamming distance between any pair of them is at least √ n − C n. Hint: use an appropriate martingale to show that more than 2/3 of the vectors are within √ distance C n/2 of S. 4. Using Janson’s Inequality, find a threshold function for the property: G(n, p) contains at least n/10 pairwise vertex disjoint copies of K5 . 5. (i) Is it true that for every ǫ > 0 there is a finite constant C = C(ǫ) such that every set X of n points in the plane contains a subset Y of size at most C with the following property: any convex set K which is the intersection of 10 half-planes and contains at least ǫn points of X, contains at least one point of Y ? Prove, or give a counterexample. (ii) Is it true that for every ǫ > 0 there is a finite constant C = C(ǫ) such that every set X of n points in the plane contains a subset Y of size at most C with the following property: any convex set K which contains at least ǫn points of X, contains at least one point of Y ? Prove, or give a counterexample.

4 libxz - dwixehpianewa zeizexazqd zehiy 317610087 l`kin bxaxtiw 2009 i`na 22

1 dl`y S

`id

mb ik xexa .(dcig` zebltzdae z"a ote`a da xag zeidl xgap xai` lky) zixwn dveaw-zz miiw

E [(Gi [S ])] = E [(Gi [S ])]

(Gi [S ]) + (Gi [S ]) > (Gi ) = k-e

oke

S

 [n] idz

li`ed .zixwn dveaw-zz

 = E [(Gi [S ])] > k=2 xear

fv j ci (v) 6 j g onqp .fgg zeivwpet zveawe Gi [S ] mitxbd oia zedfl ozip .Gi zriav ci : [n] ! [k] idz :Azuma q"r okle Lipschitz i`pz z` zniiwn L ik ze`xl lw .L = -e 0 6 j 6 k h h p p pi pi ln(2m) 2 ln(2m) k 6 Pr  (Gi [S ]) <  2 ln(2m) k < e = 1= 2 m Pr  (Gi [S ]) < k=2

Bj

=

6 i 6 m xear Gi [S ]; Gi [S ] mitxbdn cg` zegtly zexazqdd o`kn

okle ,1-n (ynn) dphw yexcd z` miiwi `l 1

.A2 =

S -e A1

=

S

xicbdl xzep .yexck

S

zniiw

2 dl`y

 4 6 0:64-y jk xgap 0 6 j 6 2n xear Bj = [j ] onqp .Si -a opeazp .zeivwpet n jezn ixwn ote`a f : [2n] ! [n] divwpet xgap 
 1 n okle Lipschitz i`pz z` zniiwn Li ik ze`xl lw .i = E [jf (Si )j] = n 1 1 ik xexa .Li (f ) = jf (Si )j-e n .jynda xgai

n0 .e

2

= 2 -y jk



xicbpe 0:63

61

1=e

2n

h Pr oekp ipyd oeieey-i`d)

0:63

jLi (f )

61

1=e

i j 6

jLi (f )

i j >

6

p

p

2 4n ln(4e n )

4 = lim

i n

p

p

4

.n0 = maxi

2 ln(4m)

2n

2 4n ln(e4 n )

i

:Azuma q"r

< 1=m

6 4 n :1 6 i 6 m lkly jk f zniiwy o`kn zrk .(milecb witqn mi-n xear

6 lim Lin(f ) 6 lim ni + 4 = 1

1=e + 4

6 0:64

fn0i g xicbp .0:63 6 Li (f ) 6 0:64 ,n > n0i lkly jk n0i miiw xnelk 3 dl`y

S -a

xehweel

g -n

:mipeieey-i`d

Hamming wgxn `ed L(g ) ,Zn3 -a xehwe zbviin g miniiw Azuma q"r okle Lipschitz i`pz z` zniiwn L ik

:

ixrfn

Pr



p



L(g ) >  +  n < e

2 =2

Pr



L(g ) < 

B

! A ,B = [n] ,A = [3] xicbp 6 i 6 n xear Bi = [i]-e

ze`xl lw .0

p



 n n

:mialwne

lim (n dveaw

V1



+
=2) 6 lim



C 00 n

!0

Cn5

1 + C 00 n C

4



  2=5 1+C 00 1=10 hxtae p  n 2=5 la`

C

= lim((C 00 1

) )= 1

C 00 n

 V (G) iaihwecpi` ote`a zeveaw ly dxcq xicbp .K5 z` dlikn n=2 lceba dveaw lk hrnk ,xnelk G[S2 ] ' K5 -y jk S2  V2 -e V2  V (G) V1 .G[S1 ] ' K5 -y jk S1  V1 -e n=2 lceba idylk .yexck ,k > n=10 ,jSi j = 5-e li`ed .jV (G) S1 [


[ Sk j > n=2

cer lk d`ld jke

` 5 dl`y mixeyin-ivgd lk ly

V C -d

cnin ik dcaera aygzdae ,Spencer -e

Alon

ly mxtq jezn 14.4.3 dpwqn itl .oekp

10 ly jezig opidy (zexenwd) zeveawd lk ly V C -d cnin ik lawp ,3 `ed xeyina zecewp lrn ( ) divwpet zniiw ,14.4.5 htynn okl .iteq hxtae ,2
3
10
log(3
10) lr dler epi` xeyina zecewp

lrn mixeyin-ivg

X

lky jk

C 

jY j 6 C mr Y

.

zyx- dlikn

a 5 dl`y ipt lr zecewp

n

> 2C ly dveaw zeidl X z` xgap

.edylk

C

.X -n zecewp

jX

Y

j>n

C

= C ()-e  = 1=2 idi .oky dlilya gipp .oekp `l

idi .jY j 6 C -y calae idylk Y  X idz .lbrn > n=2 = n likne Y -n dcewp s` likn `l K ,ok enk .xenw K ik xexa .X Y -n

zecewpd md eicewcwy (eil` zkiiy dneqgd d`tdy jk) rlevn

2

K